Solved Problems

Source Transformation

Master Network Theorems

Dr. Mithun Mondal

Tutorial 10 — Source Transformation

A practical voltage source \(V_s\) in series with a resistance \(R\) is equivalent, at its terminals, to a current source \(I_s = V_s/R\) in parallel with the same \(R\) — and vice versa. By repeatedly swapping between these two forms, series and parallel elements can be merged until the circuit collapses to a single loop or node, making the target voltage or current easy to read off. These problems apply the technique to circuits with independent voltage and current sources.

Network Theorems · Electric Circuit Analysis · 7 solved problems

Demonstrative VideoWalkthrough
Problem 1Mixed Sources

Find \(i_a\) by simplifying the circuit using source transformation.

Circuit for Problem 1
Solution

Step 1The 2 A source in parallel with the 8 Ω resistor becomes a voltage source \(2 \times 8 = 16\ \text{V}\) in series with 8 Ω:

Step 1 transformation for Problem 1

Step 2The 16 V source in series with \(8 + 4 = 12\ \Omega\) is converted back to a current source \(16/12 = 4/3\ \text{A}\) in parallel with 12 Ω, which now sits beside the 6 Ω:

Step 2 transformation for Problem 1

Step 3Combining \(6 \parallel 12 = 4\ \Omega\) and transforming the \(4/3\ \text{A}\) source with its 4 Ω back to a voltage source \(\tfrac{4}{3}\times 4 = \tfrac{16}{3}\ \text{V}\) reduces the network to a single loop:

Step 3 single-loop circuit for Problem 1

Applying KVL around the final loop:

\[ \begin{aligned} -10 + 3 i_a + 4 i_a - \tfrac{16}{3} &= 0\\[2pt] 7 i_a &= 10 + \tfrac{16}{3} = \tfrac{46}{3}\\[2pt] i_a &= \tfrac{46}{21} = 2.19\ \text{A} \end{aligned} \]
Answer\(i_a = 2.19\ \text{A}\)
Problem 2Current Divider

Find \(i_{R2}\) by simplifying the circuit using source transformation.

Circuit for Problem 2
Solution

Step 1Convert the 20 V source in series with \(R_1 = 5\ \text{k}\Omega\) into a current source. Its arrow points in the same sense as \(I_S\) (toward the top node):

\[ I_{ST} = \frac{V_s}{R_1} = \frac{20}{5} = 4\ \text{mA} \]

Step 2\(I_{ST}\) and the 2 mA source \(I_S\) are now in parallel and aid each other, so they combine into a single source:

\[ I_P = I_{ST} + I_S = 4 + 2 = 6\ \text{mA} \]

With \(I_P\) feeding \(R_1 \parallel R_2\), the current through \(R_2\) follows the current-divider rule:

\[ i_{R2} = \frac{R_1}{R_1 + R_2}\, I_P = \frac{5}{5 + 7.5}\times 6 = 2.4\ \text{mA} \]
Answer\(i_{R2} = 2.4\ \text{mA}\)
Problem 3Ladder Reduction

Find \(v_a\) using source transformation.

Circuit for Problem 3
Solution

Step 1On the right, the 30 mA source in parallel with 100 Ω becomes \(0.03 \times 100 = 3\ \text{V}\) in series with 100 Ω, which adds to the 8 V source to give an 11 V branch:

Step 1 voltage-source form for Problem 3

Step 2Convert the 10 V–100 Ω and 11 V–100 Ω branches into current sources (100 mA and 110 mA), leaving three 100 Ω resistors in parallel with the two sources across the \(v_a\) node:

Step 2 current-source form for Problem 3

Step 3The two sources add to \(210\ \text{mA}\) and two of the 100 Ω resistors combine to \(50\ \Omega\), so \(v_a\) appears across \(50 \parallel 100\):

\[ v_a = \frac{50 \times 100}{50 + 100}\times 0.21 = 7\ \text{V} \]
Answer\(v_a = 7\ \text{V}\)
Problem 4Multi-Stage Reduction

Find \(v_0\) if \(i = 5/2\ \text{A}\), using source transformation.

Circuit for Problem 4
Solution

Step 1Transform the 2 A–16 Ω pair into a 32 V source and merge series resistances on the left branch:

Step 1 reduction for Problem 4

Step 2Repeated transformations collapse the upper resistors and the 3 A source, reducing the network toward a single loop:

Step 2 reduction for Problem 4
Step 3 reduction for Problem 4

Step 3The final single loop contains a 6 V source, 9 Ω, a 36 V source, 19 Ω and \(v_0\):

Final loop for Problem 4

Applying KVL with the given \(i = 5/2\ \text{A}\):

\[ \begin{aligned} -6 + i(9 + 19) - 36 - v_0 &= 0\\[2pt] v_0 &= 28 i - 42\\[2pt] v_0 &= 28\!\left(\tfrac{5}{2}\right) - 42 = 28\ \text{V} \end{aligned} \]
Answer\(v_0 = 28\ \text{V}\)
Problem 5Single Equivalent Source

Obtain a single current source for the network shown below at terminals \(A\text{–}B\), and find the magnitude of that current source together with its parallel resistance.

Circuit for Problem 5
Solution

Convert each practical voltage source (source in series with its resistor) into its current-source form. The arrow of each equivalent source points toward the terminal nearest the source's \(+\) sign, i.e. toward node \(A\) for both branches:

\[ I_1 = \frac{18}{6} = 3\ \text{A},\qquad I_2 = \frac{10}{5} = 2\ \text{A} \]

Both sources are now in parallel and drive current in the same direction (both have their \(+\) terminal toward \(A\)), so they add:

\[ I_{eq} = I_1 + I_2 = 3 + 2 = 5\ \text{A} \]

The two parallel internal resistances \(r_1 = 6\ \Omega\) and \(r_2 = 5\ \Omega\) combine into:

\[ R = \frac{r_1\, r_2}{r_1 + r_2} = \frac{6 \times 5}{6 + 5} = \frac{30}{11} = 2.73\ \Omega \]
Current-source form for Problem 5
Note on direction. Because both 18 V and 10 V sources have their \(+\) terminals at the top (toward \(A\)), their Norton currents point the same way and the equivalent source is \(I_1 + I_2 = 5\ \text{A}\). Only if one source were reversed would the currents oppose and give \(|I_1 - I_2| = 1\ \text{A}\).
Answer\(I_{eq} = 5\ \text{A},\quad R = 2.73\ \Omega\)

Additional Practice Problems

Two self-contained worked examples with schematics: combining a voltage source with a parallel current source, and a multi-step transformation chain that collapses a network to one loop.

Problem 6Voltage→Current, Combine

A 24 V source in series with a 6 Ω resistor feeds node \(A\). A 2 A current source (also directed into \(A\)) and a 12 Ω resistor are connected from \(A\) to the reference. Find the node voltage \(V_A\) and the current through the 12 Ω resistor.

+ − 24V 6Ω A 12Ω 2A
Solution

Transform the 24 V source with its 6 Ω resistor into a current source directed into \(A\):

\[ I_{ST} = \frac{24}{6} = 4\ \text{A}\ \parallel\ 6\ \Omega \]

This 4 A source aids the existing 2 A source, and the 6 Ω now sits in parallel with the 12 Ω:

\[ I_P = 4 + 2 = 6\ \text{A},\qquad R_P = 6 \parallel 12 = \frac{6 \times 12}{18} = 4\ \Omega \]

The node voltage and the 12 Ω current follow directly:

\[ V_A = I_P\, R_P = 6 \times 4 = 24\ \text{V},\qquad I_{12} = \frac{V_A}{12} = 2\ \text{A} \]
Answer\(V_A = 24\ \text{V},\ I_{12} = 2\ \text{A}\)
Problem 7Transformation Chain

A 6 A source is in parallel with a 3 Ω resistor at node \(A\). A 3 Ω resistor connects \(A\) to node \(B\), where a 6 Ω resistor returns to the reference. Find the current \(i\) in the 6 Ω resistor and the voltage \(v_B\) across it.

6A 3Ω A 3Ω B 6Ω vₐ i
Solution

Step 1Transform the 6 A source in parallel with 3 Ω into a voltage source:

\[ V = 6 \times 3 = 18\ \text{V}\ \text{in series with}\ 3\ \Omega \]

Step 2This 3 Ω is now in series with the connecting 3 Ω, so they add, leaving an 18 V source behind 6 Ω driving the 6 Ω at \(B\):

\[ R_{series} = 3 + 3 = 6\ \Omega \]

The circuit is now a single loop, so:

\[ i = \frac{18}{6 + 6} = 1.5\ \text{A},\qquad v_B = i \times 6 = 9\ \text{V} \]
Answer\(i = 1.5\ \text{A},\ v_B = 9\ \text{V}\)
Electric Circuits & Networks